Difference between revisions of "Equivalent conditions for a linear map between two normed spaces to be continuous everywhere/1 implies 2"

Revision as of 22:35, 26 February 2016

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Needs some finishing off with the conclusion

Statement

Given two normed spaces $(X,\Vert\cdot\Vert_X)$ and $(Y,\Vert\cdot\Vert_Y)$ and also a linear map $L:X\rightarrow Y$ then we have:

If $L$ maps a sequence, $(x_n)_{n=1}^\infty\rightarrow 0$ (a null sequence) to a bounded sequence then
$L$ is continuous at some $p\in X$

Proof

This is a proof by contrapositive. That is we will show that if $L$ is not continuous at $p$ $\implies$ $L$ takes a null sequence to one that isn't bounded (an unbounded one).

Let the normed spaces X and Y be given, as well as a linear map L:X→Y
- Suppose that $L$ is not continuous at $p$ , this means:

[Expand]

$\exists (x_n)_{n=1}^\infty\rightarrow p$ such that $\lim_{n\rightarrow\infty}\left(L(x_n)\right)\ne L(p)$

Recall that continuity states that:

$L$ is continuous at $p$ $\iff$ $\forall(x_n)_{n=1}^\infty\subseteq X \left[\left(\lim_{n\rightarrow\infty}(x_n)=p\right)\implies\left(\lim_{n\rightarrow\infty}\left(L(x_n)\right)=L\left(\lim_{n\rightarrow\infty}(x_n)\right)=L(p)\right)\right]$

So it follows that to not be continuous at $p$ :

∃(xn)∞n=1⊆X[(limn→∞(xn)=p)∧¬(limn→∞(L(xn))=L(p))], by negation of implies. Additionally we may negate the = and thus we see this is the same as:
- $\exists (x_n)_{n=1}^\infty\subseteq X\left[\left(\lim_{n\rightarrow\infty}(x_n)=p\right)\wedge\left(\lim_{n\rightarrow\infty}(L(x_n))\ne L(p)\right)\right]$

Which is exactly "there exists a sequence in $X$ whose limit is $p$ and where $\lim_{n\rightarrow\infty}(L(x_n))\ne L(p)$ "

Let us now take L(xn)↛ and subtract L(p) from both sides. We see:
- L(x_n)-L(p)\not\rightarrow L(p)-L(p), using the fact that L is linear we see that:
  - $L(x_n-p)\not\rightarrow L(0)$ and $L(0)=0\in Y$ so:
- $L(x_n-p)\not\rightarrow 0$
Thus $\Vert L(x_n-p)\Vert_Y\not\rightarrow 0$ (as $\Vert0\Vert_Y=0$ by definition)

[Expand]

So $\exists C>0\ \forall N\in\mathbb{N}\ \exists n\in\mathbb{N}[n>N\wedge\Vert L(x_n-p)\Vert_Y>\epsilon]$

If a sequence converges to $0$ then we have:

$\forall\epsilon>0\ \exists N\in\mathbb{N}\ \forall n\in\mathbb{N}[n>N\implies d(x_n,x)<\epsilon]$ we know we don't have this, so we negate it:
$\exists\epsilon>0\ \forall N\in\mathbb{N}\ \exists n\in\mathbb{N}[n>N\wedge d(x_n,x)>\epsilon]$ .

That is there exists an $\epsilon>0$ such that for all $N$ there exists a bigger $n$ such that $d(x_n,x)>\epsilon$ , we shall later call such an $\epsilon$ $C$ and construct a subsequence out of the $n$ s

Thus it is possible to construct a subsequence, $(\Vert L(x_{n_k}-p)\Vert_Y)_{k=1}^\infty$ of the image $(x_n)$ where for every $k$ we have:

[Expand]

$\Vert L(x_{n_k}-p)\Vert_Y>C$

By the negation of convergent sequence we see that there exists a $C$ such that for all $N\in\mathbb{N}$ there exists another natural number, $n$ such that $\Vert L(x_n-p)\Vert_Y>C$ , we construct a sequence, $(n_k)_{k=1}^\infty$ of such $n$ -ns. That is we know there exists a $C$ , so we pick $N=1$ and get the $n$ that works, this is our first term, we then set $N=2$ and the $n$ that works is our second term, and so forth.

Some of these $n$ s may be the same, but that doesn't matter.

$(x_{n_k})$ is the subsequence of $(x_n)$ which contains only the terms that satisfy $\Vert L(x_n-p)\Vert_Y> C$

It would have been better if I used $n$ and $m$ as terms, to make which $n$ I am talking about clearer, but a reader able to attempt this proof should follow.

We now have a sequence $(x_{n_k})$ such that $\Vert L(x_{n_k}-p)\Vert_Y>C$
Define a new sequence b_k:=\frac{1}{\sqrt{\Vert x_{n_k}-p\Vert} }
- It is easy to see that $b_k\rightarrow +\infty$ (as $(x_{n_k}-p)\rightarrow 0$ )
Define a new sequence $d_k:=b_k(x_{n_k}-p)$

TODO: How does $d_k\rightarrow 0\in X$ ? Use the metric induced by the $\Vert\cdot\Vert_Y$ ?

Clearly $\Vert d_k\Vert_X=b_k\Vert x_{n_k}-p\Vert_X=\frac{\Vert x_{n_k}-p\Vert_X}{\sqrt{\Vert x_{n_k}-p\Vert_X} }$ $=\sqrt{\Vert x_{n_k}-p\Vert_X} \rightarrow 0$
But $\Vert L(d_k)\Vert_Y=\Vert L(b_k(x_{n_k}-p))\Vert_Y=b_k\Vert L(x_{n_k}-p)\Vert_Y\ge Cb_k\rightarrow +\infty$

Thus we have shown if $L$ is not continuous at $p$ that the mapping of a null sequence is unbounded, the contrapositive of what we set out to claim

@@ Line 49: / Line 49: @@
 :** It is easy to see that {{M|b_k\rightarrow +\infty}} (as {{M|(x_{n_k}-p)\rightarrow 0}})
 :* Define a new sequence {{MM|1=d_k:=b_k(x_{n_k}-p) }}
-{{Todo|How does {{M|d_k\rightarrow 0\in X}}?}}
+{{Todo|How does {{M|d_k\rightarrow 0\in X}}? Use the metric induced by the {{M|\Vert\cdot\Vert_Y}}?}}
 :* Clearly {{MM|1=\Vert d_k\Vert_X=b_k\Vert x_{n_k}-p\Vert_X=\frac{\Vert x_{n_k}-p\Vert_X}{\sqrt{\Vert x_{n_k}-p\Vert_X} } }} {{MM|1= =\sqrt{\Vert x_{n_k}-p\Vert_X} \rightarrow 0}}
 :* But {{MM|1=\Vert L(d_k)\Vert_Y=\Vert L(b_k(x_{n_k}-p))\Vert_Y=b_k\Vert L(x_{n_k}-p)\Vert_Y\ge Cb_k\rightarrow +\infty}}
-* Thus we have shown if {{M|L}} is not continuous at 0 that the mapping of a null sequence is unbounded, the [[contrapositive]] of what we set out to claim
+* Thus we have shown if {{M|L}} is not continuous at {{M|p}} that the mapping of a null sequence is unbounded, the [[contrapositive]] of what we set out to claim
 <noinclude>
 {{Theorem Of|Linear Algebra|Functional Analysis}}
 </noinclude>

Difference between revisions of "Equivalent conditions for a linear map between two normed spaces to be continuous everywhere/1 implies 2"

Revision as of 22:35, 26 February 2016

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